Posted by **Sarah** on Wednesday, October 10, 2007 at 12:39pm.

A 250.0 kg roller coaster car has 20000 J of potential energy at the top of a hill. Neglecting frictional losses, what is the velocity of the car at the bottom of the hill?

- Physical Science-please help -
**bobpursley**, Wednesday, October 10, 2007 at 12:46pm
The KE will equal the PE at the top

1/2 m v^2= 20000

solve for v.

- Physical Science-please help -
**Sarah**, Wednesday, October 10, 2007 at 1:05pm
I believe that this answer is 9.8 m/s am I right?

- Physical Science-please help -
**bobpursley**, Wednesday, October 10, 2007 at 1:29pm
The kinetic energy (1/2 *mass*velocity squared) will equal the potential energy a the top

1/2 m v^{2} = 20000

The veloicity will be near 13 meters/second at the bottom

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